1. Identify the centers of the circles:
- Let the center of $\omega_1$ be $I$.
- Let the center of $\omega_2$ be $E$.
- Note that $\omega_1$ is tangent to $DA$, $AB$, and $BC$, and $\omega_2$ is tangent to $BC$, $CD$, and $DA$.

2. Define the point of intersection:
- Let $K$ be the point of intersection of $AD$ and $BC$.

3. Understand the homothety:
- The center $E'$ of $\omega_1$ is the image of the center $E$ of the excircle of $\triangle DCK$ corresponding to the edge $DC$ through a homothety centered at $K$ with ratio $\frac{KA}{KD}$.

4. Prove similarity of triangles:
- Triangles $KDI$ and $KCE$ are similar. This can be shown by noting that both triangles share angle $\angle DKI$ and $\angle KID = \angle KEC$ because $I$ and $E$ are centers of circles tangent to the same lines.

5. Use the similarity to find angles:
- Since $KA/KD = KE'/KE$, the triangles $KIA$ and $KCE'$ are similar. Hence, $\angle KIA \equiv \angle E'CK$.

6. Calculate the angles:
- We have $\angle DAI = \frac{\angle DKI}{2} + \angle AIK = \frac{\angle DKI}{2} + \angle E'CK$.
- Therefore, $2\angle DAI = \angle DKI + 2\angle KCE'$.

7. Use the property of tangents:
- The angle between tangents from a point to a circle is twice the angle between a tangent and the line joining that point to the center of that circle.
- Hence, the angle between $BK$ and $\ell_2$ is $2\angle KCE'$, and the angle between $KD$ and $\ell_1$ is $2\angle DAI$.

8. Conclude parallelism:
- From the equation $2\angle DAI = \angle DKI + 2\angle KCE'$, we can deduce that $\ell_1$ and $\ell_2$ intersect $KC$ at the same angles, hence they are parallel.

$$\boxed{\ell_1 \parallel \ell_2}$$